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  • $\begingroup$ That was beautiful. How the heck did you come up with the idea of the recursive function? It was so simple now you showed me how you did it. Spent hours on this question and couldn't do it. Thanks! $\endgroup$ Commented May 26, 2014 at 3:40
  • $\begingroup$ You're welcome. Just keep "recursive" in mind. $\endgroup$ Commented May 26, 2014 at 3:44
  • $\begingroup$ I don't understand how to resolve the recursion n(h) from the answer. If it din't have +1, it would be the formula for the n-th Fibonacci number $\endgroup$ Commented Sep 10, 2015 at 12:30
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    $\begingroup$ @MaksimDmitriev It's better to make this recursive equation, linear homogeneous with constant coefficients (read about it: link). We must make that "+ 1" disappear. We can do this by simplifying $n(h) - n(h-1) = (n(h-1) + n(h-2) + 1) - (n(h-2) + n(h-3) + 1)$. This lead to $n(h) = 2n(h-1) - n(h-3)$. Then you can read Wikipedia article mentioned above to solve it. Note that for this new recursive equation you need 3 initial conditions ($n(0), n(1)$ and $n(2)$). $\endgroup$ Commented Sep 10, 2015 at 16:44